Integrability of Continuous Functions

From Continuity to a Small Darboux Gap

The Mesh Bound for the Darboux Gap of a Monotone Function used monotonicity to control the total change across a partition. Continuity provides a different kind of control: when two points are close, their function values are close, uniformly over the interval. This lets us bound the oscillation on every cell of a fine partition, even when the function is not monotone.

The key distinction is between continuity at each individual point and uniform continuity across the whole interval. On a closed bounded interval, the Heine–Cantor Theorem says that continuity implies uniform continuity. Once that fact is available, a partition with sufficiently small mesh has small oscillation on every cell. Summing those oscillations times the cell lengths gives a small Darboux gap.

Definition (Modulus of Continuity): Let $f:[a,b]\to\mathbb{R}$ be bounded. For $\delta>0$, define $$ \omega_f(\delta)=\sup\{|f(x)-f(y)|:x,y\in[a,b],\ |x-y|\leq\delta\}. $$ The function $\omega_f$ measures the largest possible change in $f$ between points at distance at most $\delta$. We say that $f$ has a modulus tending to zero if $\omega_f(\delta)\to0$ as $\delta\to0^+$.

The supremum in this definition is finite because $f$ is bounded. The set over which it is taken is nonempty, since it includes pairs with $x=y$. The modulus is nondecreasing: if $0<\delta_1\leq\delta_2$, then every pair allowed for $\delta_1$ is also allowed for $\delta_2$, so $\omega_f(\delta_1)\leq\omega_f(\delta_2)$.

A bounded function is uniformly continuous exactly when its modulus tends to zero. For one direction, given $\varepsilon>0$, uniform continuity supplies $\eta>0$ such that $|f(x)-f(y)|<\varepsilon/2$ whenever $|x-y|<\eta$. If $0<\delta<\eta$, every pair used to define $\omega_f(\delta)$ has distance less than $\eta$, and therefore $\omega_f(\delta)\leq\varepsilon/2<\varepsilon$. Conversely, if the modulus tends to zero, choose $\delta>0$ such that $\omega_f(\delta)<\varepsilon$. Then $|x-y|<\delta$ implies $|f(x)-f(y)|\leq\omega_f(\delta)<\varepsilon$, which is uniform continuity.

A General Darboux-Gap Estimate

For a bounded function, the oscillation on a cell is its supremum there minus its infimum there. If the cell has length at most $\delta$, any two points in it are within distance $\delta$. The modulus therefore bounds the oscillation on that cell. The Darboux gap is the sum of the cell oscillations weighted by their lengths.

Theorem (Darboux-Gap Bound from a Modulus of Continuity): Let $f:[a,b]\to\mathbb{R}$ be bounded, and let $P=\{x_0,\ldots,x_n\}$ be a partition of $[a,b]$. For every $\delta>0$ such that $\|P\|\leq\delta$, $$ 0\leq U(f,P)-L(f,P)\leq (b-a)\omega_f(\delta). $$

Proof. For the $i$th cell $[x_{i-1},x_i]$, write $M_i=\sup f([x_{i-1},x_i])$ and $m_i=\inf f([x_{i-1},x_i])$. If $u,v$ belong to this cell, then $|u-v|\leq x_i-x_{i-1}\leq\|P\|\leq\delta$, so $|f(u)-f(v)|\leq\omega_f(\delta)$. It follows that $M_i-m_i\leq\omega_f(\delta)$: indeed, the bound holds for every pair of values $f(u),f(v)$, and taking the supremum over one value and the infimum over the other preserves the inequality.

By the Oscillation Formula for the Darboux Gap,

\begin{aligned} U(f,P)-L(f,P) &=\sum_{i=1}^{n}(x_i-x_{i-1})(M_i-m_i)\\ &\leq\sum_{i=1}^{n}(x_i-x_{i-1})\omega_f(\delta)\\ &=(b-a)\omega_f(\delta). \end{aligned}

The gap is nonnegative because each cell supremum is at least its infimum. This proves both inequalities. $\square$

This estimate separates the two sources of error: the interval contributes its total length $b-a$, while the function contributes its largest change over distances up to $\delta$. If the modulus tends to zero, reducing the mesh makes the Darboux gap arbitrarily small.

Continuous Functions on Closed Intervals

Theorem (Integrability of Continuous Functions): If $f:[a,b]\to\mathbb{R}$ is continuous and $a<b$, then $f$ is Riemann integrable on $[a,b]$.

Proof. By the Extreme Value Theorem, $f$ is bounded on $[a,b]$. By the Heine–Cantor Theorem, it is uniformly continuous there. Thus its modulus of continuity tends to zero, as shown above.

Let $\varepsilon>0$. Since $\omega_f(\delta)\to0$, choose $\delta>0$ such that $$ (b-a)\omega_f(\delta)<\varepsilon. $$ There exists a partition $P$ of $[a,b]$ with $\|P\|<\delta$, by the proposition that uniform partitions have arbitrarily small mesh. The Darboux-Gap Bound from a Modulus of Continuity now gives

0\leq U(f,P)-L(f,P) \leq(b-a)\omega_f(\delta) <\varepsilon.

The Darboux Criterion says that a bounded function is Riemann integrable if for every positive tolerance there is a partition whose Darboux gap is smaller than that tolerance. We have established boundedness and constructed such a partition for each $\varepsilon>0$. The criterion proves that $f$ is Riemann integrable. $\square$

The proof uses continuity in two distinct ways: boundedness ensures that upper and lower sums are defined, and uniform continuity makes their gap small. Neither part can simply be omitted from this argument. For a function on a general, noncompact interval, continuity alone need not give boundedness or uniform continuity.

Worked Examples with Explicit Mesh Bounds

Worked Example: A Quadratic on a Symmetric Interval

Let $f(x)=x^2$ on $[-1,1]$. For $x,y\in[-1,1]$,

|f(x)-f(y)|=|x^2-y^2|=|x-y||x+y|\leq2|x-y|,

because $|x+y|\leq|x|+|y|\leq2$. Thus $\omega_f(\delta)\leq2\delta$. If $P$ is a uniform partition into $n$ cells, then its mesh is $2/n$, and the Darboux-gap bound gives

U(f,P)-L(f,P)\leq 2\cdot\omega_f(2/n) \leq2\cdot\frac{4}{n}=\frac{8}{n}.

For example, $n=81$ gives a gap at most $8/81<0.1$, since $8<8.1$. This certifies integrability using only a bound on changes in function values; it does not require computing the upper or lower sums exactly.

Worked Example: The Square-Root Function

Let $g(x)=\sqrt{x}$ on $[0,1]$. For $x,y\geq0$, we may assume $x\geq y$ when estimating the absolute difference. Then

(\sqrt{x}-\sqrt{y})^2=x+y-2\sqrt{xy}\leq x-y,

because the inequality is equivalent to $y\leq\sqrt{xy}$, which follows from $y^2\leq xy$. Therefore $|\sqrt{x}-\sqrt{y}|\leq\sqrt{x-y}$; reversing the roles of $x$ and $y$ gives $$ |\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}. $$ So $\omega_g(\delta)\leq\sqrt{\delta}$. A uniform partition into $n$ cells has mesh $1/n$, and hence

U(g,P)-L(g,P)\leq1\cdot\sqrt{1/n}=\frac{1}{\sqrt{n}}.

For a gap below $0.01$, it suffices to take $n>10{,}000$, because then $1/\sqrt n<0.01$. The function is uniformly continuous, but it is not Lipschitz on $[0,1]$. If it had a Lipschitz constant $L$, taking $x=1/n$ and $y=0$ would give

\frac{1}{\sqrt n} =|\sqrt{1/n}-\sqrt0| \leq L|1/n-0| =\frac{L}{n},

and therefore $\sqrt n\leq L$ for every positive integer $n$. This is impossible because $\sqrt n$ is unbounded. The direct contradiction shows that the square-root modulus does not amount to a Lipschitz bound.

Worked Example: An Absolute-Value Function with a Corner

Let $h(x)=|x-1|$ on $[0,3]$. The reverse triangle inequality gives

|h(x)-h(y)| =\bigl||x-1|-|y-1|\bigr| \leq |(x-1)-(y-1)| =|x-y|.

Thus $\omega_h(\delta)\leq\delta$. For a uniform partition into $n$ cells, the mesh is $3/n$, so

U(h,P)-L(h,P)\leq3\cdot\frac{3}{n}=\frac{9}{n}.

Taking $n=91$ gives a gap at most $9/91<0.1$, since $90<91$. The change in slope at $x=1$ causes no problem: the estimate applies across the corner as well as on either side of it.

Why the Modulus Viewpoint Is Useful

The Darboux Criterion is qualitative: it asks for arbitrarily small gaps. A modulus of continuity makes the argument quantitative by telling us how small the mesh should be. If a bound $\omega_f(\delta)\leq\phi(\delta)$ is known, where $\phi(\delta)\to0$, it is enough to choose a partition with mesh at most $\delta$ such that $(b-a)\phi(\delta)<\varepsilon$. Every such partition then has Darboux gap below $\varepsilon$.

The square-root example illustrates why it is useful not to require a Lipschitz estimate. A Lipschitz bound has the form $|f(x)-f(y)|\leq L|x-y|$, which yields a gap bound proportional to the mesh. Uniform continuity asks only that changes in function values become small as the distance becomes small; the resulting rate can be slower, as $\sqrt{\delta}$ is for the square-root function. That slower rate still suffices for integrability.

A common error is to confuse continuity at every point with uniform continuity without checking the domain. The Heine–Cantor Theorem applies here because the domain is the closed bounded interval $[a,b]$. Another is to infer that a function is Lipschitz merely because it is uniformly continuous; the square-root calculation disproves that implication. The integrability proof needs uniform continuity, not the stronger Lipschitz property.

Check boundedness.
For a continuous function on a closed interval, the Extreme Value Theorem supplies it.

Control changes over short distances.
Use uniform continuity, or obtain an explicit modulus bound.

Choose a small mesh.
Make the product of interval length and the modulus bound smaller than the desired tolerance.

Apply the Darboux Criterion.
The small Darboux gap, together with boundedness, proves integrability.

Check Your Understanding

Use the modulus and Darboux-gap estimates to answer the following questions.

Why is the modulus of continuity finite for a bounded function on $[a,b]$?
If $\omega_f(\delta)\leq 3\delta$ on $[0,2]$, what upper bound does the gap estimate give for a partition of mesh at most $\delta$?
Which two properties of a continuous function on a closed interval are used separately in the integrability proof?
For $g(x)=\sqrt{x}$ on $[0,1]$, what mesh bound ensures a Darboux gap below $\varepsilon$?
Why does uniform continuity not imply that a function is Lipschitz?